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# I need some help with these Algebra problems. Thank you in advance for your time and effort in helping me.

1.  √x(√9-4/3)=√25
The sqrt is just over the 9.

2.  (4-3*√2)2

3.  √135b2c3d*√5b2d
The sqrt is over 135b2c3d and 5b2d

You should include all argument under radical inside a parenthesis.

### 2 Answers by Expert Tutors

Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (7 lesson ratings) (7)
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1) √x(√9-[4/3])=√25
√x(3-[4/3])=5
√x(5/3)=5
√x(5/3)(3/5)=5(3/5)
√x=3
x=9

3) (√135b2c3d)*(√5b2d)
3bc(√15cd)*b(√5d)
3bc(√5d√3c)*b(√5d)
(3bc)(b)(√5d)(√5d)(√3c)
(3b2c)(5d)(√3c)
15b2cd√3c

Arthur:
4th line:

√(X . 5/3) = 5
then:
X(5/3) = 25
X = 25. 3/5 = 15

You forgot to square the 5 ( right side).

Parviz,
I thought the first √ was only for x, not for what is in the parentheses. This problem is misleading because we don't know for sure what the radicand is for that very first radical sign.  Your first comment says it all; this makes me think you too aren't really sure what is under that first radical sign:the x only or the x and everything in parentheses. Also, the way I solved it, I multiplied 5 by 3/5 to get 3 and then squared the 3.
I guess the only one who knows for sure is the student and so far there is no feedback. We'll just have to keep trying to figure out exactly what the problem is and take it from there.
Arthur D.
Yes, Student's writing is not very clear. They should inscribe radical's argument inside parenthesis.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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1. √X(√9 - 4/3 ) = √25

Square both sides of the equation:

X ( 3 - 4/3 ) = 25

X = 25 / ( 3 - 4/3 )

X = 25 /( 5/3)

X = 75/5 = 15

2.
( 4 - 3 √2 ) = 16 - 24 √2  + 18 = 34 - 24 √2      ( Use of the relation of ( a ±b)2 = a2 + b2 ±2ab

3.
√( 135b2 c3 d * √(5b2 d ) =

3bc √(15 cd )  * b √5d ) =

15b2 d c2√(5c)